Electromagnetic theory miscellaneous
- The velocity of the plane wave esin2(ωt – βx) is—
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The wave is
esin2(ωt − βx) = esin2β ω t − x t
When the argument is reduced to the form kt – x, the coefficient of t is the velocity of propagation. Hence, the velocity of the wave is ω/β.Correct Option: D
The wave is
esin2(ωt − βx) = esin2β ω t − x t
When the argument is reduced to the form kt – x, the coefficient of t is the velocity of propagation. Hence, the velocity of the wave is ω/β.
- The frequency of the power wave associated with an electromagnetic wave having an E field as
E = e– z/ δ cos (ωt – z/δ),
is given by—
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E = e– z/δ cos (ωt – z/δ)
The radian frequency of E wave is ω, same would be the radian frequency of associated H wave. The frequency of power wave is double the corresponding frequency of E or H wave. Thus the radian frequency is 2 ω and cyclic frequency 2ω/ 2π = ω/ π.Correct Option: D
E = e– z/δ cos (ωt – z/δ)
The radian frequency of E wave is ω, same would be the radian frequency of associated H wave. The frequency of power wave is double the corresponding frequency of E or H wave. Thus the radian frequency is 2 ω and cyclic frequency 2ω/ 2π = ω/ π.
- A plane electromagnetic wave in free space is specified by the electric field
E = ax [20 cos (ωt – βz) + 5 cos (ωt + βz)] V/m.
The associated magnetic field is—
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E and H are related by η
i.e. E/ H = η,
the characteristic impedance of the medium which is 120 π for free space. Also, phase of H in the reflected component of the wave has to reverse so that travelling of the reflected wave is correctly given by the direction of E × H.
Of course, E × H are perpendicular to each other in the incident component or the forward travelling component of the wave. Keeping in view the above facts and the given expression for E.
H = ay (1/ 120 π) 20 cos (ωt – βz) – 5 cos (ωt + βz) A /m.Correct Option: C
E and H are related by η
i.e. E/ H = η,
the characteristic impedance of the medium which is 120 π for free space. Also, phase of H in the reflected component of the wave has to reverse so that travelling of the reflected wave is correctly given by the direction of E × H.
Of course, E × H are perpendicular to each other in the incident component or the forward travelling component of the wave. Keeping in view the above facts and the given expression for E.
H = ay (1/ 120 π) 20 cos (ωt – βz) – 5 cos (ωt + βz) A /m.
- The maximum usable frequency of an ionospheric layer at 60º incidance and with 8 MHz. The critical frequency is—
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Maximum usable frequency, muf = fc sec θ
where, given θ i = 60º
fc = 8 MHz
Now, muf = 8 × 106 × sec 60º
= 16 MHzCorrect Option: A
Maximum usable frequency, muf = fc sec θ
where, given θ i = 60º
fc = 8 MHz
Now, muf = 8 × 106 × sec 60º
= 16 MHz
- The potential difference between the forces A and B of a uniformly polarized infinite slab shown in the figure is—
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D = ∈0 E + P also D = ∈0∈r E
∴∈0∈r E = ∈0 E + P or E = P ∈0(∈r − 1)
∴Potential difference V = Ed = Pd ∈0(∈r − 1) Correct Option: A
D = ∈0 E + P also D = ∈0∈r E
∴∈0∈r E = ∈0 E + P or E = P ∈0(∈r − 1)
∴Potential difference V = Ed = Pd ∈0(∈r − 1)